Question

Find the common factors of the given terms: \[14pq,28{p^2}{q^2}\]
(A) \[14pq\]
(B) \[14{p^2}q\]
(C) \[14p{q^2}\]
(D) \[pq\]

Answer 1

Hint: According to the question, firstly we will find the factors of \[14pq\] and \[28{p^2}{q^2}\] separately. Hence, find the common factors between \[14pq\] and \[28{p^2}{q^2}\] to get the required answer.

Complete step-by-step answer:
 First we will start with making the factors of \[14pq\] with the help of factorization method.
Here, we will start dividing the number with the smallest prime number that is 2 till it leaves quotient 1.
So, we will start with the given value that is 14pq.
$\Rightarrow$ 14pq = \[\dfrac{{14pq}}{2}\]
On dividing with 2 we get = 7pq
Here, again we will continue dividing the quotient with the next prime number 7 that is 7pq.
$\Rightarrow$ 7pq = \[\dfrac{{7pq}}{7}\]
On dividing with 7 we get = pq
Here, again we will continue dividing the quotient with p that is pq.
$\Rightarrow$ pq = \[\dfrac{{pq}}{p}\]
On dividing with p we get = q
Here, again we will continue dividing the quotient with q that is q.
$\Rightarrow$ q = \[\dfrac{q}{q}\]
On dividing with q we get = 1 which is our required quotient.
So, factors of \[14pq = 2 \times 7 \times p \times q\]
Secondly, we will start with making the factors of \[28{p^2}{q^2}\] with the help of factorization method.
Here, we will start dividing the number with the smallest prime number that is 2 till it leaves quotient 1.
So, we will start with the given value that is \[28{p^2}{q^2}\].
$\Rightarrow$ \[28{p^2}{q^2}\] = \[\dfrac{{28{p^2}{q^2}}}{2}\]
On dividing with 2 we get = \[14{p^2}{q^2}\]
Here, again we will continue dividing with 2 that is \[14{p^2}{q^2}\]
$\Rightarrow$ \[14{p^2}{q^2}\] = \[\dfrac{{14{p^2}{q^2}}}{2}\]
On dividing with 2 we get = \[7{p^2}{q^2}\]
Here, again we will continue dividing the quotient with next prime number 7 that is \[7{p^2}{q^2}\]
$\Rightarrow$ \[7{p^2}{q^2}\] = \[\dfrac{{7{p^2}{q^2}}}{7}\]
On dividing with 7 we get = \[{p^2}{q^2}\]
Here, again we will continue dividing the quotient with p that is \[{p^2}{q^2}\].
$\Rightarrow$ \[{p^2}{q^2}\] = \[\dfrac{{{p^2}{q^2}}}{p}\]
On dividing with p we get = \[p{q^2}\]
Here, again we will continue dividing the quotient with p that is \[p{q^2}\].
$\Rightarrow$ \[p{q^2}\] = \[\dfrac{{p{q^2}}}{p}\]
On dividing with p we get = \[{q^2}\]
Here, again we will continue dividing the quotient with q that is \[{q^2}\] .
$\Rightarrow$ \[{q^2}\] = \[\dfrac{{{q^2}}}{q}\]
On dividing with q we get = \[q\]
Here, again we will continue dividing the quotient with q that is \[q\] .
$\Rightarrow$ \[q\] = \[\dfrac{q}{q}\]
On dividing with q we get = 1 which is our required quotient.
So, factors of \[28{p^2}{q^2} = 1 \times 2 \times 7 \times p \times p \times q \times q\]
Now, we will find the common factors of \[14pq\] and \[28{p^2}{q^2}\] .
So, the common numbers between them is:
\[ \Rightarrow 2 \times 7 \times p \times q\]
After multiplying we get,
\[ \Rightarrow 14pq\]
So, option (A) \[14pq\] is correct.

Note: To solve these types of questions, we should not get confused between the factors and the multiples of a number. At last do not forget to multiply all the common factors to get the desired answer.